Why Does \(\frac{1}{i}\) Equal Both \(i\) and \(-i\)?

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The discussion clarifies that \(\frac{1}{i}\) equals both \(i\) and \(-i\) due to the properties of square roots in complex numbers. By multiplying \(\frac{1}{i}\) by \(\frac{i}{i}\), it simplifies to \(-i\). However, when squaring both sides, the equation \((\frac{1}{i})^2 = -1\) reveals that both \(i\) and \(-i\) are valid solutions, emphasizing the necessity of considering both roots in complex arithmetic.

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latentcorpse
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i've confused myself

[itex]\frac{1}{i}=-i[/itex] by multiplying by i/i

but then [itex](\frac{1}{i})^2=-1 \Rightarrow \frac{1}{i}= i,-i[/itex]
and clearly the positive root here doesn't make any sense?
 
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You are correct but you forgot to take the negative root of the left hand side.
 
Well you've somewhat answered your own question. You simply did pick the incorrect root as there are always 2 roots to a square root

[tex]\[<br /> \begin{array}{l}<br /> 3 = 3 \\ <br /> 3^2 = 3^2 \\ <br /> 9 = 9 \\ <br /> \sqrt 9 = \sqrt 9 \\ <br /> 3 = - 3? \\ <br /> \end{array}<br /> \][/tex]

for example.
 

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